Question

Решить уравнение 6sin^2x+5cosx-2=0 , принадлежащие отрезку [2П;7П/2]

Answer 1

$6\sin^2x+5\cos x-2=0\\6(1-\cos^2x)+5\cos x-2=0\\6-6\cos^2x+5\cos x-2=0\\-6\cos^2x+5\cos x+4=0\;\;\;\;\;\times(-1)\\6\cos^2x-5\cos x-4=0\\\cos x=t,\;\cos^2x=t^2,\;t\in[-1;\;1]\\\\6t^2-5t-4=0\\D=25-4\cdot6\cdot(-4)=25+96-121\\t_{1,2}=\frac{5\pm11}{12}\\t_1=-\frac12\\t_2=\frac43>1\;-\;He\;nogx.\\\cos x=-\frac12\Rightarrow \boxed{x=\pm\frac{2\pi}3+2\pi n,\;n\in\mathbb{Z}}\\\\\\x\in\left[2\pi;\;\frac{7\pi}2\right]:\\\\2\pi\leq\pm\frac{2\pi}3+2\pi n\leq\frac{7\pi}2$

$\begin{array}{ccc}2\pi\leq\frac{2\pi}3+2\pi n\leq\frac{7\pi}2&|&2\pi\leq-\frac{2\pi}3+2\pi n\leq\frac{7\pi}2\\&|&\\2\pi-\frac{2\pi}3\leq2\pi n\leq\frac{7\pi}2-\frac{2\pi}3&|&2\pi+\frac{2\pi}3\leq2\pi n\leq\frac{7\pi}2+\frac{2\pi}3\\&|&\\\frac{4\pi}3\leq2\pi n\leq\frac{17\pi}6&|&\frac{8\pi}3\leq2\pi n\leq\frac{25\pi}6\\&|&\\\frac23\leq n\leq\frac{17}{12}&|&\frac43\leq n\leq\frac{25}{12}\\&|&\\n=1&|&n=2\\&|&\\\boxed{x_1=\frac{2\pi}3+2\pi=\frac{8\pi}3}&|&\boxed{x_2=-\frac{2\pi}3+4\pi=\frac{10\pi}3}\end{array}$